Linear equation in sine and cosine via double-angle

Problem

Given the equation $\;\sin^{2}x+2\sin x\cos x-\cos^{2}x-2=0$, rewrite it in linear form in sine and cosine of the variable $t=2x$ and solve it.

Solution

Use $2\sin x\cos x=\sin 2x$ and $\sin^{2}x-\cos^{2}x=-\cos 2x$. The equation becomes $-\cos 2x+\sin 2x-2=0\ \Longrightarrow\ \sin t-\cos t=2\quad(t=2x).$ Writing $\sin t-\cos t=\sqrt2\,\sin\!\left(t-\tfrac{\pi}{4}\right)$, this would require $\sqrt2\,\sin\!\left(t-\tfrac{\pi}{4}\right)=2$, i.e. $\sin\!\left(t-\tfrac{\pi}{4}\right)=\sqrt2>1$: impossible. The equation has no solutions. $\boxed{\,\nexists\,x\in\mathbb{R}\,}$